The domain of the function f(x)=1{sinx}+{sin(π+x)} (where {.} denotes the fractional part) is

f(x)= 1{sinx}+{sin(π+x)}=1{sinx}+{−sinx } is defined iff{sinx}+{−sinx} ≠0⇒ sinx≠ interger⇒ sinx≠ ± 1, 0⇒ x≠ nπ2Domain = R−{ nπ2: n is an integer}

The domain of the function f(x)=1{sinx}+{sin(π+x)} (where {.} denotes the fractional part) is

f(x)= 1{sinx}+{sin(π+x)}=1{sinx}+{−sinx } is defined iff{sinx}+{−sinx} ≠0⇒ sinx≠ interger⇒ sinx≠ ± 1, 0⇒ x≠ nπ2Domain = R−{ nπ2: n is an integer}

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$T h e d o m a i n o f t h e f u n c t i o n f (x) = \frac{1}{\sqrt{{\sin x} + {\sin (π + x)}}} (w h e r e {.} d e n o t e s t h e f r a c t i o n a l p a r t) i s$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(0, π)$

$[0, π]$

$R - {\frac{n π}{2} : n i s a n i n t e g e r}$

${(2 n + 1) \frac{π}{2} : n i s a n i n t e g e r}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{\sqrt{{\sin x} + {\sin (π + x)}}} \\ = \frac{1}{\sqrt{{\sin x} + {- sinx}}} i s d e f i n e d i f f \\ {\sin x} + {- \sin x} \neq 0 \\ \Rightarrow \sin x \neq int e r g e r \\ \Rightarrow \sin x \neq \pm 1, 0 \\ \Rightarrow x \neq \frac{n π}{2} \\ D o m a i n = R - {\frac{n π}{2} : n i s a n int e g e r} \end{array}$